Tidal Force — Physica

Equivalent forms

stretch along axis

\Delta a_{\parallel} = +\frac{2GMR}{d^{3}}

squeeze perpendicular

\Delta a_{\perp} = -\frac{GMR}{d^{3}}

tidal potential

\Phi_{t} = -\frac{GMR^{2}}{2d^{3}}\left(3\cos^{2}\theta - 1\right)

Take the difference of two inverse-square pulls and an entire third power appears: tides are the derivative of gravity.

Symbol	Name	SI unit	Dimension	Range
$F_{\text{tidal}}$	tidal force across the bodyoutput	N	M L T^{-2}	—
$G$	gravitational constant	6.674\times10^{-11} N·m²/kg²	M^{-1} L^{3} T^{-2}	—
$M$	mass of the tide-raising body (Moon, Sun)	kg	M	—
$m$	mass of the affected parcel	kg	M	—
$R$	radius of the affected body	m	L	—
$d$	distance between the two bodies' centers	m	L	—

Discovery

Isaac Newton · 1687

In the Principia, Newton was the first to explain why there are two high tides a day — a fact that had baffled everyone since antiquity (Galileo got it wrong, blaming Earth's spin). Newton showed the ocean facing the Moon is pulled away from the Earth, and the Earth is pulled away from the far-side ocean, so the water bulges on BOTH sides. He even estimated the Sun's tide from the observed spring–neap cycle.

Real-world applications

Ocean tides: lunar bulge $\approx 0.54\,\mathrm{m}$ equilibrium height, solar $\approx 0.25\,\mathrm{m}$ ; aligned at new/full moon they add (spring tides), at quarter moons they partly cancel (neap tides).
Tidal locking: the Moon's own tidal bulge braked its spin until one face always points at Earth.
Earth's slowing rotation: tidal friction lengthens the $day \sim 2.3\,\mathrm{ms}$ per century, and the Moon recedes 3.8 cm/yr.
Io's volcanoes: Jupiter's tides knead the moon's interior, making it the most volcanic world known.
Spaghettification near black holes — the same $1/d^{3}$ stretch taken to the extreme.
Roche limit: inside $\sim 2.4$ planetary radii, tides exceed a moon's self-gravity and tear it apart — that's where Saturn's rings live.

Common misconceptions

“The Moon lifts the water on one side only.” — There are two bulges; the far side bulges because the Earth itself is pulled away from that water.
“The Sun's gravity on Earth is weaker than the Moon's, that's why lunar tides win.” — The Sun's pull $is \sim 180\times$ stronger! But tides go as $M/d^{3}$ , and the Moon's closeness wins by a factor $\sim 2.2$ .
“High tide is directly under the Moon.” — Friction and continents delay and distort the bulge by hours.

Experimental verification

Global tide-gauge networks and satellite altimetry (TOPEX/Poseidon, Jason) map the ocean tide to millimeter accuracy; lunar laser ranging confirms the Moon's 3.8 cm/yr recession predicted by tidal momentum transfer.

Derivation

Acceleration of a point at distance $d - R$ from the Moon: a_near $= GM/(d-R)^{2}$ .
Acceleration of Earth's center: a_ $c = GM/d^{2}$ .
Difference: $\Delta a = GM[1/(d-R)^{2} - 1/d^{2}]$ .
Expand with R ≪ d using $(1-x)^-2 \approx 1 + 2x$ : $\Delta a \approx (GM/d^{2})(2R/d) = 2GMR/d^{3}$ .
Multiplying by the parcel mass m gives F_tidal $\approx 2$ GMmR/ $d^{3}$ along the Moon–Earth axis; the same expansion sideways gives a squeeze half as strong.

Limiting cases

R ≪ d⟶

F \approx 2

GMmR/

d^{3}

The standard tidal approximation used everywhere from oceans to binary stars.

d \to

Roche limit⟶ tides ≈ self-gravitySatellites inside this distance are torn apart — the origin of planetary rings.

What if…

What if the Moon were twice as close?

Tides go as $1/d^{3}$ , so they'd be $8\times$ higher — coastal cities would flood twice a day under several meters of water.

What if Earth had no Moon?

Only solar tides $\sim 45$ % of today's lunar ones. Days would also be shorter — lunar tidal friction is what braked Earth from $a \sim 6$ -hour spin to 24 hours.

1

Moon vs Sun: who raises the bigger tide?

Given · M_moon

= 7.35\times 10^{22} kg

, d_moon

= 3.84\times 10^{8} m

;

M_sun = 1.99\times 10^{30} kg

,

d_sun = 1.496\times 10^{11} m

Find · Ratio of lunar to solar tidal force on Earth

Steps

Tidal force $\propto M/d^{3}$ .
Moon: $M/d^{3} = 7.35\times 10^{22} / (3.84\times 10^{8})^{3} = 1.30\times 10^{-3} kg/m^{3}$ .
Sun: $M/d^{3} = 1.99\times 10^{30} / (1.496\times 10^{11})^{3} = 5.94\times 10^{-4} kg/m^{3}$ .
Ratio $= 1.30\times 10^{-3} / 5.94\times 10^{-4} \approx 2.2$ .

Result · The Moon's tide

is \approx 2.2\times the Sun's

— even though the Sun pulls on Earth

\sim 180\times

harder overall.

2

Tidal acceleration across an astronaut near a stellar black hole

Given ·

M = 10

solar masses

= 1.99\times 10^{31} kg

,

d = 100\,\mathrm{km}

, astronaut length

R = 1\,\mathrm{m}

(head-to-toe half-length

\approx 1\,\mathrm{m})

Find · Differential acceleration head-to-toe

Steps

$\Delta a = 2GMR/d^{3} = 2 \times 6.674\times 10^{-11} \times 1.99\times 10^{31} \times 1 / (10^{5})^{3}$
$\Delta a = 2.66\times 10^{21} / 10^{15} = 2.66\times 10^{6} m/s^{2}$ .

Result ·

\approx 2.7

million g across one meter — spaghettification long before the horizon.

Newton's Law of Universal Gravitation→Kepler's Third Law→Gravitational Potential Energy→Escape Velocity→